Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

Last updated at April 19, 2021 by Teachoo

Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

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Misc 15 Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4π/3 . Given, Radius of sphere = r Let R be the radius of the cone and H be its height. Let β BOC = ΞΈ Now, AC = AO + OC H = r + r cos ΞΈ H = r (1 + cos ΞΈ) And, R = r sin ΞΈ We need to maximize volume of cone. Volume of the cone is V = 1/3 ππ ^2 π» V = 1/3 ππ^2 sin^2β‘"ΞΈ" r (1 + cos ΞΈ) V = π/π π π^π γπππγ^πβ‘"ΞΈ" (1 + cos ΞΈ) Finding π π½/π π½ ππ/ππ = 1/3 ππ^3 (π[sin^2β‘γΞΈ γ (1 + cosβ‘ΞΈ))/ππ ππ/ππ = 1/3 ππ^3 [2 sinβ‘γ"ΞΈ" cosβ‘γ"ΞΈ" (1+cosβ‘γ"ΞΈ" )+sin^2β‘γ"ΞΈ" (βsinβ‘γ"ΞΈ" )γ γ γ γ γ ] ππ/ππ = 1/3 ππ^3 (2 sinβ‘γ"ΞΈ" cosβ‘γ"ΞΈ" (1+cosβ‘γ"ΞΈ" )βsin^3β‘"ΞΈ" γ γ γ ) ππ/ππ = 1/3 ππ^3 (2 sinβ‘γ"ΞΈ" cosβ‘γ"ΞΈ" (1+cosβ‘γ"ΞΈ" )βsin^3β‘"ΞΈ" γ γ γ ) ππ/ππ = 1/3 ππ^3 sin ΞΈ (2 cosβ‘γ"ΞΈ" +γ2 cos^2γβ‘γ"ΞΈ" βγπππγ^πβ‘"ΞΈ" γ γ ) ππ/ππ = 1/3 ππ^3 sin ΞΈ (2 cosβ‘γ"ΞΈ" +γ2 cos^2γβ‘γ"ΞΈ" β(πβγπππγ^πβ‘"ΞΈ" γ γ)) ππ/ππ = 1/3 ππ^3 sin ΞΈ (π γπππγ^πβ‘γ"ΞΈ" +γπ πππγβ‘γ"ΞΈ" βπγ γ ) ππ/ππ = 1/3 ππ^3 sin ΞΈ (π γπππγ^πβ‘γ"ΞΈ" +γπ πππγβ‘π½ γ βπππβ‘π½ βπ) ππ/ππ = 1/3 ππ^3 sin ΞΈ (3 γπππ γβ‘"ΞΈ" (cosβ‘π+1)β1(cosβ‘π+1)) ππ/ππ = 1/3 ππ^3 sin ΞΈ (3 πππ β‘γ"ΞΈ" β1γ )(πππ β‘γ"ΞΈ" +1γ ) Putting π π½/π π½ = 0 1/3 ππ^3 sin ΞΈ (πππ β‘γ"ΞΈ" +1γ ) (3 πππ β‘γ"ΞΈ" β1γ ) = 0 π¬π’π§ ΞΈ (cos ΞΈ + 1) (3 cos "ΞΈ" β 1) = 0 π¬π’π§ ΞΈ = 0 ΞΈ = 0Β° ΞΈ cannot be 0Β° for cone ππ¨π¬ ΞΈ + 1 = 0 cos ΞΈ = β1 For cone, 0Β° < ΞΈ < 90Β° & cos ΞΈ is negative in 2nd & 3rd quadrant. So cos ΞΈ = β1 is not possible π ππ¨π¬ ΞΈ β 1 = 0 cos ΞΈ = 1/3 ΞΈ = cosβ1 1/3 cos ΞΈ = π/π is possible So, πππβ‘"ΞΈ" = π/π Thus, H = r (1 + cos ΞΈ) H = r ("1 + " 1/3) H = ππ/π Finding (π ^π π½)/(π π½^π ) ππ/ππ = (1/3 ππ^3 sinβ‘π (3 cos2 π + 2 cosβ‘πβ 1)) = 1/3 ππ^3 [cosβ‘π (3 cos^2β‘π+2 cosβ‘πβ1)+sinβ‘π (6 cosβ‘π (βπ ππ)β2 sinβ‘π )] = 1/3 ππ^3 [3 cos^3β‘π+2 cos^2β‘πβcosβ‘π+sinβ‘π (β6 cosβ‘π sinβ‘πβ2 sinβ‘π )] = 1/3 ππ^3 [3 cos^2β‘π+2 cos^2β‘πβcosβ‘πβ6 sin^2β‘π cosβ‘πβ2 sin^2β‘π ] Now, ππ¨π¬β‘π½=π/π And sin2 "ΞΈ" = 1 β cos2 "ΞΈ" = 1 β (1/3)^2= 1 β 1/9 = π/π Putting values in (π ^π π½)/(π π½^π ) (π^2 π)/(ππ^2 ) = 1/3 ππ^3 [3(1/3)^3+2(1/3)^2β1/3β6(1/3)(8/9)β2(8/9)] = 1/3 ππ^3 [3(1/27)+2(1/9)β1/3β6(8/27)β2(8/9)] = 1/3 ππ^3 [1/9+ 2/9β1/3β16/9β16/9] = 1/3 ππ^3 [(β32)/9] = (βπππ π^π)/ππ Thus, (π^2 π)/(ππ^2 ) < 0 for ππ¨π¬β‘π½=π/π So, V is maximum for ππ¨π¬β‘π½=π/π Hence, altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is ππ/π .

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Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.